Question: In right triangle $PQR$, we have $\angle Q = \angle R$ and $QR = 6\sqrt{2}$.  What is the area of $\triangle PQR$?
A triangle can't have two right angles, so a right triangle with two congruent angles must have congruent acute angles.  That is, $\triangle PQR$ must be an isosceles right triangle with acute angles at $Q$ and $R$.  Therefore, $\overline{QR}$ is the hypotenuse of the triangle, and $QP=RP=\frac{QR}{\sqrt{2}}$, which means $QP=RP=6$ and $[QRP]=(QP)(RP)/2 = \boxed{18}$.

[asy]
unitsize(1inch);
pair P,Q,R;
P = (0,0);
Q= (1,0);
R = (0,1);
draw (P--Q--R--P,linewidth(0.9));
draw(rightanglemark(Q,P,R,3));
label("$P$",P,S);
label("$Q$",Q,S);
label("$R$",R,N);
[/asy]